65. Valid Number LeetCode Solution

In this guide, you will get 65. Valid Number LeetCode Solution with the best time and space complexity. The solution to Valid Number problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.

Problem Statement
Complexity Analysis
Valid Number solution in C++
Valid Number solution in Java
Valid Number solution in Python
Additional Resources

65. Valid Number LeetCode Solution image

Problem Statement of Valid Number

Given a string s, return whether s is a valid number.

For example, all the following are valid numbers: “2”, “0089”, “-0.1”, “+3.14”, “4.”, “-.9”, “2e10”, “-90E3”, “3e+7”, “+6e-1”, “53.5e93”, “-123.456e789”, while the following are not valid numbers: “abc”, “1a”, “1e”, “e3”, “99e2.5”, “–6”, “-+3”, “95a54e53”.
Formally, a valid number is defined using one of the following definitions:

An integer number followed by an optional exponent.
A decimal number followed by an optional exponent.

An integer number is defined with an optional sign ‘-‘ or ‘+’ followed by digits.
A decimal number is defined with an optional sign ‘-‘ or ‘+’ followed by one of the following definitions:

Digits followed by a dot ‘.’.
Digits followed by a dot ‘.’ followed by digits.
A dot ‘.’ followed by digits.

An exponent is defined with an exponent notation ‘e’ or ‘E’ followed by an integer number.
The digits are defined as one or more digits.

Example 1:

Input: s = “0”
Output: true

Example 2:

Input: s = “e”
Output: false

Example 3:

Input: s = “.”
Output: false

Constraints:

1 <= s.length <= 20
s consists of only English letters (both uppercase and lowercase), digits (0-9), plus '+', minus '-', or dot '.'.

Complexity Analysis

Time Complexity: O(n)
Space Complexity: O(1)

65. Valid Number LeetCode Solution in C++

class Solution {
 public:
  bool isNumber(string s) {
    trim(s);
    if (s.empty())
      return false;

    bool seenNum = false;
    bool seenDot = false;
    bool seenE = false;

    for (int i = 0; i < s.length(); ++i) {
      switch (s[i]) {
        case '.':
          if (seenDot || seenE)
            return false;
          seenDot = true;
          break;
        case 'e':
        case 'E':
          if (seenE || !seenNum)
            return false;
          seenE = true;
          seenNum = false;
          break;
        case '+':
        case '-':
          if (i > 0 && s[i - 1] != 'e' && s[i - 1] != 'E')
            return false;
          seenNum = false;
          break;
        default:
          if (!isdigit(s[i]))
            return false;
          seenNum = true;
      }
    }

    return seenNum;
  }

 private:
  void trim(string& s) {
    s.erase(0, s.find_first_not_of(' '));
    s.erase(s.find_last_not_of(' ') + 1);
  }
};
/* code provided by PROGIEZ */

65. Valid Number LeetCode Solution in Java

class Solution {
  public boolean isNumber(String s) {
    s = s.trim();
    if (s.isEmpty())
      return false;

    boolean seenNum = false;
    boolean seenDot = false;
    boolean seenE = false;

    for (int i = 0; i < s.length(); ++i) {
      switch (s.charAt(i)) {
        case '.':
          if (seenDot || seenE)
            return false;
          seenDot = true;
          break;
        case 'e':
        case 'E':
          if (seenE || !seenNum)
            return false;
          seenE = true;
          seenNum = false;
          break;
        case '+':
        case '-':
          if (i > 0 && s.charAt(i - 1) != 'e' && s.charAt(i - 1) != 'E')
            return false;
          seenNum = false;
          break;
        default:
          if (!Character.isDigit(s.charAt(i)))
            return false;
          seenNum = true;
      }
    }

    return seenNum;
  }
}
// code provided by PROGIEZ

65. Valid Number LeetCode Solution in Python

class Solution:
  def isNumber(self, s: str) -> bool:
    s = s.strip()
    if not s:
      return False

    seenNum = False
    seenDot = False
    seenE = False

    for i, c in enumerate(s):
      if c == '.':
        if seenDot or seenE:
          return False
        seenDot = True
      elif c == 'e' or c == 'E':
        if seenE or not seenNum:
          return False
        seenE = True
        seenNum = False
      elif c in '+-':
        if i > 0 and s[i - 1] not in 'eE':
          return False
        seenNum = False
      else:
        if not c.isdigit():
          return False
        seenNum = True

    return seenNum
 # code by PROGIEZ